How do we show that on a parabolic Hölder space, the polynomial and Hölder seminorms are equivalent?

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I am currently working through Krylov's Lectures on Elliptic and Parabolic Equations in Holder Spaces. One of the key points of chapter 8 is that the two seminorms

$$[u]_{1+\delta/2,2+\delta;U} := \sup_{X \neq Y \in U}\frac{|u(X)-u(Y)|}{\rho(X,Y)^{\delta}}$$ and $$[u]'_{1+\delta/2,2+\delta;U} := \sup_{X\in Q}\sup_{\rho>0} \frac{1}{\rho^{2+\delta}}\inf_{p\in\mathcal{P}_2}|u-p|_{0;U}$$ for polynomials $p \in \mathcal{P}_2$ (those that are first order in space and second order in time) are equivalent. Here, $$\rho(X=(t,x),Y=(s,y)) := |t-s|^{1/2} + |x-y|$$ and $$Q_{\rho}(X_0) := (t_0-\rho^2) \times B_{\rho}(X_0).$$

The key point of the proof that is absolutely stumping me is that after defining the finite difference operator $$ \sigma_{h}: u\to \frac{1}{h^2}[u(t,x)-u(t-h^2,x)]$$ and showing that for any polynomial $p \in \mathcal{P}_2$ and points $X_1,X_2$ such that $t_1 \leq t_2$, taking $h := \varepsilon\rho$ for $\varepsilon \in (0,1)$ yields that \begin{equation} \tag{1}|\sigma_h(u-p)(x_i)| \leq \frac{4}{h^2}|u-p|_{0;Q_{3\rho(X_2)}}. \end{equation} Moreover we can conclude that there exists some constant $N$ such that

$$ \tag{2}\frac{4}{h^2}|u-p|_{0;Q_{3\rho(X_2)}} \leq N\varepsilon^{-2}\rho^{\delta}[u]'_{1+\delta/2,2+\delta}. $$

My main question is, how do you go from (1) to (2)? I've tried approximating the order of the differences using a Taylor polynomial, but my bound is dependent on p and u, not just the dimension of the space $d$. I've tried removing infimums and supremums, but the fact that $p$ is a specific polynomial and $[u]'$ infimizes over all polynomials seems to get me the opposite inequality.

The content of this question seems hyperspecific. If I am missing any key details I will attempt to edit them in, but I believe I have put everything relevant in to the best of my ability.

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Putting the brain to grindstone yielded what I believe is a correct solution. We note that the inequality above holds for all $p \in \mathcal{P}_2$. Therefore, taking an infimum over $p$ of both sides of the inequality yields

$$\inf_{p}\frac{4}{h^2}|u-p|_{0;Q_{3\rho(X_2)}} = \inf_{p}\frac{4}{\varepsilon^2}\cdot\frac{\rho^{\delta}}{\rho^{2+\delta}}|u-p|_{0;Q_{3\rho(X_2)}}.$$ Noting the independence of some quantities on $p$ yields $$\frac{4\rho^{\delta}}{\varepsilon^2}\frac{1}{\rho^{2+\delta}}\inf_p|u-p|_{0;Q_{3\rho(X_2)}} \leq C\varepsilon^{-2}\rho^{\delta}[u]'_{1+\delta/2,2+\delta}.$$